Every third integer is a multiple of 3: 1,2,3,4,5,6,7,8,9....

Any time you multiply integers together and one of them is a multiple of 3, the product will be divisible by 3.

x and x-1 are consecutive integers. Because they are factors of this expression, it will certainly be divisible by three if either one of them is a multiple of 3, like (6,7) or (9,10).

The only way it could fail to be divisible by three is if x and x-1 fall between the multiples of three, like if they were (7,8). In that case making the expression divisible by three would fall entirely on the shoulders of the third factor, (x-k). In this example, x=8 and x-1=7. Plugging in values of k from the answer choices with x=8 gives us 12, 10, 9, 6, 3 for A,B,C,D,E, respectively. B is the only choice for k that fails to make x-k a multiple of three, so it is the only choice of k that would fail to GUARANTEE that the product is ALWAYS divisible by 3.

If you don't like plugging in numbers, consider the following sequence:

x-5,x-4,x-3,x-2,x-1,x,x+1,x+2,x+3,x+4

Again, if x-1 and x are not multiples of 3, then x+1, x+4, x-2, and x-5 must be. These would correspond to A,C,D,E for x-k

GmatMathPro wrote:Every third integer is a multiple of 3: 1,2,3,4,5,6,7,8,9....

Any time you multiply integers together and one of them is a multiple of 3, the product will be divisible by 3.

x and x-1 are consecutive integers. Because they are factors of this expression, it will certainly be divisible by three if either one of them is a multiple of 3, like (6,7) or (9,10).

The only way it could fail to be divisible by three is if x and x-1 fall between the multiples of three, like if they were (7,8). In that case making the expression divisible by three would fall entirely on the shoulders of the third factor, (x-k). In this example, x=8 and x-1=7. Plugging in values of k from the answer choices with x=8 gives us 12, 10, 9, 6, 3 for A,B,C,D,E, respectively. B is the only choice for k that fails to make x-k a multiple of three, so it is the only choice of k that would fail to GUARANTEE that the product is ALWAYS divisible by 3.

If you don't like plugging in numbers, consider the following sequence:

x-5,x-4,x-3,x-2,x-1,x,x+1,x+2,x+3,x+4

Again, if x-1 and x are not multiples of 3, then x+1, x+4, x-2, and x-5 must be. These would correspond to A,C,D,E for x-k

But, the question says, evenly divisible... So, does that mean, we have to look for even divisors(3*2, 3*4, 3*6..etc)

In this context, "evenly" doesn't have anything to do with even integers. When you say x divides y evenly, it just means y divided by x has no remainder. For example, 5 divides 15 evenly, because 15/3=5, even though none of those are even numbers.

This is a test of you knowledge of number properties: the product of a number of consecutive integers is divisible by its number of terms. it works to make a table listing x , x+1 and x+2 where it turns into this:

x-6, x-5, x-4
x-3, x-2, x-1
x, x+1, x+2
x+3, x+4, x+5

Thus, you need a number that fulfills the middle column, the only one that does not is B